Pluto's distance $P(t)$ (in billions of kilometers) from the sun as a function of time $t$ (in years) can be modeled by a sinusoidal expression of the form $a\cdot\sin(b\cdot t)+d$. At year $t=0$, Pluto is at its average distance from the sun, which is $6.9$ billion kilometers. In $66$ years, it is at its closest point to the sun, which is $4.4$ billion kilometers away. Find $P(t)$. $\textit{t}$ should be in radians. $P(t) = $
Explanation: The strategy First, we should convert the given information about the real-world context into mathematical terms of the sinusoidal function and its graph. Then, we should use the given information to find the amplitude, midline, and period of the function's graph. Finally, we should find $a$, $b$, and $d$ in the expression $a\sin(b\cdot t)+d$ by considering the features we found. Converting the given information into mathematical terms At $t=0$, the Pluto is $6.9$ billion kilometers from the sun. This means the graph of the function passes through $(0,6.9)$. We are given that this is the average distance, which corresponds to the midline of the graph. $66$ years later (which means $t=66$ ) the distance is $4.4$ billion kilometers. This corresponds to the point $(66,4.4)$. We are given that this is the closest point to the sun, which corresponds to a minimum point of the graph. In conclusion, the graph intersects its midline at $(0,6.9)$ and then has a minimum point at $(66,4.4)$. Determining the amplitude, midline, and period The midline intersection is at $y={6.9}$, so this is the midline. The minimum point is $2.5$ units below the midline, so the amplitude is ${2.5}$. The minimum point is $66$ units to the right of the midline intersection, so the period is $4\cdot 66={264}$. [Why did we multiply by 4?] Determining the parameters in $a\sin(b\cdot t)+d$ Since the midline intersection at $t=0$ is followed by a minimum point, we know that $a<0$. [How do we know that?] The amplitude is ${2.5}$, so $|a|={2.5}$. Since $a<0$, we can conclude that $a=-2.5$. The midline is $y={6.9}$, so $d=6.9$. The period is ${264}$, so $b=\dfrac{2\pi}{{264}}=\dfrac{\pi}{132}$. The answer $P(t)=-2.5\sin\left(\dfrac{\pi}{132}t\right)+6.9$